The function f has domain (-∞, 0) and is defines as f(x) = (x^2 + 2)/(x^2 + 5) (here ^ is used to represent a power). Show that f'(x) < 0. What is the range of f?

First notice that f(x) = u/v. So f'(x) =[ v(u') - u(v')]/v² (the Quotient rule). After working it out, we find f'(x) = 6x/(x² + 5)² (the steps can be shown on the whiteboard). Since the denominator is always positive and the numerator is always negative we conclude that f'(x) is always negative.The range of f is (2/5, 1). One way of explaining this is that when x gets very close to -∞, x² gets close to +∞ and therefore f(x) gets close to 1. When x is close to 0 (but still in the domain), the x-squared terms are very small so f(x) gets close to 2/5.